Numerical Simulations of the Oscillating Second-Grade Fluid through a Rectangular Cross Duct with Fractional Constitution Relationship

Abstract

An oscillating second-grade fluid through a rectangular cross duct is studied. A traditional integer time derivative in the kinematic tensors is substituted by a fractional operator that considers the memory characteristics. To treat the fractional governing equation, an analytical method was obtained. To analyze the impact of the parameters more intuitively, the difference method was applied to determine the numerical expression and draw with the help of computer simulation. To reduce the cost of the amount of computation and storage, a fast scheme was proposed, one which can greatly improve the calculation speed. To verify the correctness of the difference scheme, the contrast between the numerical expression and the exact expression—constructed by introducing a source term—was given and the superiority of the fast scheme is discussed. Furthermore, the influences of the involved parameters, including the parameter of retardation time, fractional parameter, magnetic parameter, and oscillatory frequency parameter, on the distributions of velocity and shear force at the wall surface with oscillatory flow are analyzed in detail.

1. Introduction

The flow of fluid has widespread applications, including in aerospace, biomedicine, oil exploitation etc. The classical fluid model is the Newtonian fluid in which the stress tensor and the kinematic tensor have a linear relationship. It has a limitation in so far that it can only describe most pure liquids such as water and alcohol. In addition to the fluids listed, most fluids are non-Newtonian whose characteristics have many properties that different from those of Newtonian ones [1]. Studying the flow mechanism has great significance. There are many types of non-Newtonian fluids and this paper studies the second-grade fluid [2,3,4], in which the shear force is characterized by the stretching tensor and the Rivlin–Ericksen tensors.

Due to the special description of the constitution relationship, the second-grade fluid has its own unique properties. In order to better discover its flow mechanism, the usual method is to consider the flow through simple models. The common categories for this include the flow on semi-infinite plates [5,6], two parallel infinitely long plates [7], the flow in pipes or ducts [8], or the flow in a circular tube [9]. Non-Newtonian fluids in rectangular channels have gained special interest for the engineering applications such as in magnetohydrodynamic generators and marine mechanical equipment, interest which has helped us to study the flow characteristics in depth [10]. Studying second-grade fluid in a rectangular cross duct has important research significance and application value. Erdoğan and İmrak [11] were the first scholars to study the unsteady motion of second-grade fluid through a rectangular cross duct with the influences of the side walls. It has been further studied by many scholars. Considering heat transfer with relaxation time, Alamri et al. [12] analyzed particle diffusion in the flow of second-grade fluid and discussed the effects of the involved parameters on the profile graphically. Bernard [13] studied a three-dimensional second-grade fluid with a tangential boundary condition in a polyhedron. By comparing with the stress of the Newtonian fluid at the initial time, Erdoğan and İmrak [14] considered the motion properties of second-grade fluid driven by the impulsive motion or sudden pressure gradient. The comparison of the stress at the start time between the Newtonian fluid and the second-grade fluid was discussed. Furthermore, the influence of the magnetic field has important research significance. It has been applied to the Maxwell fluid [15], Oldroyd-B fluid [16] et al., but it has fewer studies on the second-grade fluid.

Besides, many situations consider the steady-state motion of the second-grade fluid for simplicity. However, for the practical situation, the velocity field produced by the flow should vary with time due to the complexity of the fluid flow. The unsteady state has more research significance for the second-grade fluid with the condition that the time derivative in the constitutional relationship is integral to considering the local characteristics. With further research, it has been found that the fractional model has gained support for its memory characteristic [17]. At present, the fractional operators have been applied in many viscoelastic fluids, such as the Maxwell model [18], Oldroyd-B model [19], Burgers’ model [20] et al. For the fractional second-grade fluid, the constitution relationship has a similar form with the viscoelastic fluid, namely, they all have the fractional material derivative term. The application of fractional operator on the motion of second-grade fluid has been analyzed by Tan and Xu [21], Bazhlekova et al. [22], Kan and Wang [23], Li et al. [24] et al. For flow driven by a special form of oscillatory pressure, it has been widely applied in the motions in an isosceles right triangle tube with Maxwell fluid [25], in a straight rectangular duct with the second-grade fluid [26], in a cylindrical domain with the Oldroyd-B fluid [27], and in cylindrical domains with the fractional Burgers fluid [28]. To the best of the authors’ knowledge, the two-dimensional flow of second-grade fluid in rectangular ducts driven by oscillatory pressure and considering a magnetic field has not been considered in the literature so far.

There are many methods to solve the governing equation [29,30,31]. For the treatment of the fractional second-grade fluid, the traditional method is to apply the integral transform method to obtain an analytical solution [27,32,33], with the paradox that the principle of causality causing the initial conditions is a non-rigorous enforcement. In other words, these treatments for the start-up flow proposed by Christov [34,35] are incorrect. There are many numerical methods [36] that can solve the fractional governing equation and the numerical difference method has been applied to solve the corresponding mathematical problem correctly.

The governing equation subject to the fractional second-grade constitutive relationship is solved numerically. The difference is that the integer term has mature calculation methods, while the key is to treat the fractional derivative. The classical method is to choose the L1 scheme [37] to approximate it, though it is limited by the huge amount of computation and storage required for long-term numerical simulation, since the Caputo derivative depends on historical information. This is an urgent problem to be solved at present. Using exponential functions to approximate the Abel kernel function of Caputo derivatives, the fast algorithm [38] has been developed. The main idea is to reduce the number of iterations by constructing a recurrent relationship. At each time step, the convolution containing the exponential kernel is calculated in $\mathrm{O}\left(1\right)$ time. Then the computational amount $\mathrm{O}\left(N^2\right)$ and storage $\mathrm{O}\left(N\right)$ for the direct L1-algorithm reduces to $\mathrm{O}\left(N{\log }_{2}N\right)$ and $\mathrm{O}\left({\log }_{2}N\right)$ for the fast algorithm, respectively. This has been applied to treat fractional diffusion models [39], multi-term fractional sub-diffusion models [40], wave models [41] and the variable coefficient fractional diffusion wave models [42]. According to the numerical results, the analyses are discussed and are detailed by graphical illustration.

The paper’s outline is given as follows. The derivation of the mathematical model of the second-grade fluid over a rectangular duct with an infinite length and which is caused by a various pressure gradient is given in Section 2. The exact expression for describing the second-grade fluid is deduced in Section 3. Section 4 gives the numerical difference scheme of the formulated governing equation and the analyses of the solvability, stability and convergence are proven in Section 5. Section 6 gives the fast evolution of the difference scheme. Section 7 gives the comparison between the numerical expression and the exact expression. Furthermore, the influences of the relevant parameters on the transfer mechanism of the velocity field and the shear force at the wall surface are also analyzed. The conclusions are summarized in Section 8.

2. The Derivation of the Mathematical Model

Consider the motions of an incompressible second-grade fluid. The laminar flow in a straight duct with infinite length and the rectangular cross-section is considered and the flow is controlled by pressure gradient with time/space oscillations. As shown in Figure 1, the width and height of the rectangular section are $2a$ and $2b$. The center in the cross-section is defined as the origin and the boundaries along $x$ direction and $y$ direction are at the positions $x=\pm a$ and $y=\pm b$ while $z\in \left[0,+\infty \right)$ along $z$ direction. For simplicity, the body forces are neglected in this paper.

The continuity equation is given as

$$\nabla \cdot V=0,$$

(1)

where $\nabla$ denotes the gradient operator.

As a development, for the fractional second-grade fluid when considering the memory characteristics [21], the stress tensor $\tau$ has the following expression

$$\tau =\mu A_1+{\alpha }_1{A}_2+{\alpha }_2{A}_1^2,$$

(2)

where $\mu$ refers to the dynamic viscosity, ${\alpha }_1$ and ${\alpha }_2$ denote the material moduli, $A_1$ and $A_2$ are the kinematic tensors with the expression as

$$A_1=\nabla V+{\left(\nabla V\right)}^T \mathrm{and} A_2=D_t^{\alpha }{A}_1+A_1\nabla V+{\left(\nabla V\right)}^T{A}_1,$$

(3)

where $D_t^{\alpha }$ denotes the Riemann–Liouville’s fractional operator of order $\alpha$ ($0<a<1$) [43], the definition for a function $f(t)$ defined on $\left[t_1,t_2\right]$ is given as

$$D_t^{\alpha }f\left(t\right)=\frac{d}{dt}\left(\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle {\int }_{t_1}^t\frac{f\left(\xi \right)}{{\left(t-\xi \right)}^{\alpha }}d\xi }\right).$$

(4)

For considering the Clausius–Duhem inequality and assuming that the minimum at equilibrium for the Helmholtz free energy is [44,45], the material constants satisfy the following restrictions

$$\mu \ge 0, {\alpha }_1\ge 0 \mathrm{and} {\alpha }_1+{\alpha }_2=0.$$

(5)

Applying the periodic pressure gradient into the $z$-direction, the motion of second-grade fluid in the direction is parallel to the axial coordinate with an oscillating form. The velocity field is assumed as

$$V={\left[0,0,w\left(x,y,t\right)\right]}^T,$$

(6)

where $w\left(x,y,t\right)$ refers to the velocity in the $z$-direction. For this consideration, it is simple to find that the continuity Equation (1) automatically satisfies consideration of the velocity field (6).

Considering the effect of an electromagnetic field, the motion equation for describing the second-grade fluid is denoted as follows

$$\rho D_{t}V=-\nabla p+\nabla \cdot \tau -{\sigma }_0{B}_0^{2}V,$$

(7)

where $V$ corresponds to the velocity vector, $\rho$ refers to the fluid density, $p$ denotes the hydrostatic pressure, the operator $D_t$ refers to the material derivative, ${\sigma }_0$ refers to electrical conductivity and $B_0$ is the magnetic field.

Combining the expansion of Equation (2) (see the Appendix A) with Equation (7), the fractional governing equation can be derived as

$$\frac{\partial w}{\partial t}=\nu \left(1+\lambda D_t^{\alpha }\right)\left(\frac{{\partial }^{2}w}{\partial x^2}+\frac{{\partial }^{2}w}{\partial y^2}\right)-Mw-\frac{1}{\rho }\frac{\partial p}{\partial z},$$

(8)

where $\nu =\frac{\mu }{\rho }$ denotes the kinematic viscosity, $\lambda =\frac{{\alpha }_1}{\mu }$ refers to the retardation time, $M=\frac{{\sigma }_0{B}_0^2}{\rho }$ corresponds to the magnetic parameter.

The initial conditions are

$$w\left(x,y,0\right)=0,$$

(9)

and the boundary conditions regardless of slip are given as

$$w\left(\pm a,y,t\right)=w\left(x,\pm b,t\right)=0.$$

(10)

The initial boundary conditions of (9) and (10) are the Dirichlet type based on the physical backgrounds considering a laminar flow in a straight duct with infinite length and rectangular cross-section. When the boundary conditions change to Neumann, Robin or some other kind of initial boundary conditions, the physical meaning of this paper changes. However, the treatment process of the fractional governing equation with the difference method and the fast algorithm is also applicable. The only difference is the boundary discretization is slightly different.

Theorem 1.

[43] Assume a positive$\alpha$satisfies$0\le n-1<\alpha <n$. Suppose the function$f\left(t\right)$in region$\left[t_1,t_2\right]$has$n-1$continuous bounded derivative for every$t_2>t_1$, then

$$D_t^{\alpha }f\left(t\right)={}^{C}D_t^{\alpha }f\left(t\right)+{\displaystyle \sum _{j=1}^{n-1}\frac{f^{\left(j\right)}\left(t_1\right){\left(t-t_1\right)}^{j-\alpha }}{\mathsf{\Gamma }\left(1+j-\alpha \right)}}, t_1\le t\le t_2,$$

(11)

where${}^{C}D_t^{\alpha }f\left(t\right)$refers to the Caputo’s fractional derivative [43].

Through Theorem 1, we have $D_t^{\alpha }f\left(t\right)={}^{C}D_t^{\alpha }f\left(t\right)$ with the condition that second-grade fluid flowing along a straight rectangular duct is subjected to the zero initial condition. In the following discussions, we are able to substitute the Riemann–Liouville derivative with Caputo’s derivative.

3. Analytical Solution

In this part, we try to obtain the analytical solution of (8)–(10). Firstly, we consider the equation:

$$\frac{\partial u}{\partial t}=\nu \left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}\right)-Mu,$$

(12)

$$u\left(\pm 1,y,t,\tau \right)=u\left(x,\pm 1,t,\tau \right)=0.$$

(13)

To simplify the calculation, we introduce $u\left(x,y,t,\tau \right)=h\left(x+1,y+1,t,\tau \right)$, after which it becomes:

$$\frac{\partial h}{\partial t}=\nu \left(\frac{{\partial }^{2}h}{\partial x^2}+\frac{{\partial }^{2}h}{\partial y^2}\right)-Mh,$$

(14)

$$h\left(0,y,t,\tau \right)=h\left(x,0,t,\tau \right)=h\left(2,y,t,\tau \right)=h\left(x,2,t,\tau \right)=0$$

(15)

The solution of (14)–(15) is obtained by separation of variables. Defining $h\left(x,y,t,\tau \right)=T\left(t,\tau \right)\mathsf{\Phi }\left(x,y\right)$ and the operator $\mathsf{\Delta }=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}$, yields:

$$\mathsf{\Phi }\frac{\partial T}{\partial t}=\nu T\mathsf{\Delta }\mathsf{\Phi }-MT\mathsf{\Phi },$$

(16)

$$\mathsf{\Phi }\left(0,y\right)=\mathsf{\Phi }\left(x,0\right)=\mathsf{\Phi }\left(2,y\right)=\mathsf{\Phi }\left(x,2\right)=0$$

(17)

Denote

$$\mathsf{\Delta }\mathsf{\Phi }=\eta \cdot \mathsf{\Phi }.$$

(18)

It can be deduced immediately that

$$\frac{\partial T}{\partial t}=\left(\nu \eta -M\right)T$$

(19)

Equation (18) is a Helmholtz equation and the solution with the boundary conditions (17) can be obtained: ${\mathsf{\Phi }}_{n,m}=\sin \left(\frac{n\pi }{2}x\right)\sin \left(\frac{m\pi }{2}y\right)$, where $n,m\in \mathbb{N}$. It can then be deduced that $\eta$ can only be discrete values with the value $\eta =-\frac{n^2+m^2}{4}{\pi }^2$. It is simple to determine the solution to Equation (19) as $T=B\left(\tau \right)e^{\left(\nu \eta -M\right)t}$, where $B\left(\tau \right)$ is an arbitrary function. The solution of $h(x,y,t,\tau )$ has the following form

$$h\left(x,y,t,\tau \right)={\displaystyle \sum _{n=1}^{+\infty }{\displaystyle \sum _{m=1}^{+\infty }{B}_{n,m}\left(\tau \right)e^{-\left(v{\pi }^2\frac{n^2+m^2}{4}+M\right)t}\sin \left(\frac{n\pi }{2}x\right)\sin \left(\frac{m\pi }{2}y\right)}},$$

(20)

and then

$$u(x,y,t,\tau )={\displaystyle \sum _{n=1}^{+\infty }{\displaystyle \sum _{m=1}^{+\infty }{B}_{n,m}\left(\tau \right)e^{-\left(v{\pi }^2\frac{n^2+m^2}{4}+M\right)t}\sin \left(\frac{n\pi }{2}\left(x+1\right)\right)\sin \left(\frac{m\pi }{2}\left(y+1\right)\right)}}$$

(21)

Denote $\frac{1}{\rho }\frac{\partial p}{\partial z}=g\left(t\right)$. Equation (8) can be expressed as:

$$\frac{\partial w}{\partial t}=\nu \left(1+\lambda D_t^{\alpha }\right)\left(\frac{{\partial }^{2}w}{\partial x^2}+\frac{{\partial }^{2}w}{\partial y^2}\right)-Mw-g\left(t\right).$$

(22)

Suppose there is a function $u\left(x,y,t,\tau \right)$ satisfying $w\left(x,y,t\right)={\displaystyle {\int }_0^{t}u\left(x,y,t,\tau \right)d\tau }$. Substituting this expression into (21), yields

$$\begin{array}{l}u\left(x,y,t,t\right)+{\displaystyle {\int }_0^t\frac{\partial u\left(x,y,t,\tau \right)}{\partial t}d\tau }={\displaystyle {\int }_0^t\left(\nu \mathsf{\Delta }u\left(x,y,t,\tau \right)-Mu\left(x,y,t,\tau \right)\right)d\tau }\\ +\frac{\nu \lambda }{\mathsf{\Gamma }\left(1-\alpha \right)}\frac{d}{dt}{\displaystyle {\int }_0^t{\left(t-\xi \right)}^{-\alpha }}{\displaystyle {\int }_0^{\xi }\mathsf{\Delta }u\left(x,y,\xi ,\tau \right)d\tau d\xi }-g\left(t\right).\end{array}$$

(23)

From Equation (12), Equation (23) can be reduced as:

$$\frac{\nu \lambda }{\mathsf{\Gamma }\left(1-\alpha \right)}\frac{d}{dt}{\displaystyle {\int }_0^t{\left(t-\xi \right)}^{-\alpha }}{\displaystyle {\int }_0^{\xi }\mathsf{\Delta }u\left(x,y,\xi ,\tau \right)d\tau }d\xi -u\left(x,y,t,t\right)=g\left(t\right).$$

(24)

Substituting the solution (21) into Equation (24), yields

$$\begin{array}{l}[-\frac{\nu \lambda }{\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle \sum _{n=1}^{+\infty }{\displaystyle \sum _{m=1}^{+\infty }\left(\frac{n^2+m^2}{4}{\pi }^2\right)\frac{d}{dt}{\displaystyle {\int }_0^t{\left(t-\xi \right)}^{-\alpha }}{\displaystyle {\int }_0^{\xi }{B}_{n,m}\left(\tau \right)d\tau }{e}^{-\left(v{\pi }^2\frac{n^2+m^2}{4}+M\right)\xi }d\xi }}\\ -{\displaystyle \sum _{n=1}^{+\infty }{\displaystyle \sum _{m=1}^{+\infty }{B}_{n,m}\left(t\right)e^{-\left(v{\pi }^2\frac{n^2+m^2}{4}+M\right)t}}}]\cdot \sin \left(\frac{n\pi }{2}\left(x+1\right)\right)\sin \left(\frac{m\pi }{2}\left(y+1\right)\right)=g\left(t\right).\end{array}$$

(25)

Perform the inner product with $\sin \left(\frac{n_0\pi }{2}\left(x+1\right)\right)\sin \left(\frac{m_0\pi }{2}\left(y+1\right)\right)$ on both sides of Equation (25) and integral interval chosen as $\left[-1,1\right]\times \left[-1,1\right]$. Then for the left side of the Equation (25), we have:

$$\begin{array}{l}{\displaystyle \sum _{n=1}^{+\infty }{\displaystyle \sum _{m=1}^{+\infty }{C}_{n,m}\left(t\right){\displaystyle {\int }_{-1}^1{\displaystyle {\int }_{-1}^1\sin \left(\frac{n\pi }{2}\left(x+1\right)\right)\sin \left(\frac{m\pi }{2}\left(y+1\right)\right)\sin \left(\frac{n_0\pi }{2}\left(x+1\right)\right)\sin \left(\frac{m_0\pi }{2}\left(y+1\right)\right)dx}dy}}}\\ ={\displaystyle \sum _{n=1}^{+\infty }{\displaystyle \sum _{m=1}^{+\infty }\frac{C_{n,m}\left(t\right)}{4}}}{\displaystyle {\int }_{-1}^1\left[\cos \left(\frac{\left(n-n_0\right)\pi }{2}\left(x+1\right)\right)-\cos \left(\frac{\left(n+n_0\right)\pi }{2}\left(x+1\right)\right)\right]}dx\\ \hspace{1em}\cdot {\displaystyle {\int }_{-1}^1\left[\cos \left(\frac{\left(m-m_0\right)\pi }{2}\left(y+1\right)\right)-\cos \left(\frac{\left(m+m_0\right)\pi }{2}\left(y+1\right)\right)\right]dy}\\ =C_{n_0,m_0}\left(t\right)\end{array}$$

(26)

where

$$C_{n,m}\left(t\right)=-\frac{\nu \lambda {\pi }^2\left(n^2+m^2\right)}{4\mathsf{\Gamma }\left(1-\alpha \right)}\frac{d}{dt}{\displaystyle {\int }_0^t{\left(t-\xi \right)}^{-\alpha }}{\displaystyle {\int }_0^{\xi }{B}_{n,m}\left(\tau \right)d\tau }{e}^{-\left(v{\pi }^2\frac{n^2+m^2}{4}+M\right)\xi }d\xi -B_{n,m}\left(t\right)e^{-\left(v{\pi }^2\frac{n^2+m^2}{4}+M\right)t}$$

For the right-hand component of Equation (25), the integral is zero when $n_0$ and $m_0$ are even. Set $n_0=2k_1-1$, $m_0=2k_2-1$, where $k_1$ and $k_2$ are positive integers. Then we have the following integral formula:

$${\displaystyle {\int }_{-1}^1{\displaystyle {\int }_{-1}^1\sin \left(\frac{\left(2k_1-1\right)\pi }{2}\left(x+1\right)\right)\sin \left(\frac{\left(2k_2-1\right)\pi }{2}\left(y+1\right)\right)}}dxdy=\frac{16}{\left(2k_1-1\right)\left(2k_2-1\right){\pi }^2}.$$

(27)

By a combination of Equations (26) and (27), the following equation can be obtained:

$$\frac{C_{n,m}^{(1)}}{\mathsf{\Gamma }\left(1-\alpha \right)}\frac{d}{dt}{\displaystyle {\int }_0^t{\left(t-\xi \right)}^{-\alpha }}{\displaystyle {\int }_0^{\xi }{B}_{n,m}\left(\tau \right)d\tau }{e}^{-C_{n,m}^{(2)}\xi }d\xi -B_{n,m}\left(t\right)e^{-C_{n,m}^{(2)}t}=C_{n,m}^{(3)}g\left(t\right),$$

(28)

where $n=2k_1-1$, $m=2k_2-1$, $C_{n,m}^{(1)}=-\left(\frac{n^2+m^2}{4}{\pi }^2\right)\nu \lambda$, $C_{n,m}^{(2)}=v{\pi }^2\frac{n^2+m^2}{4}+M$ and $C_{n,m}^{(3)}=\frac{16}{nm{\pi }^2}$.

Denote $\xi =t-\gamma$, Equation (28) can be rewritten as:

$$\begin{array}{l}\frac{C_{n,m}^{(1)}}{\mathsf{\Gamma }\left(1-\alpha \right)}\left({\displaystyle {\int }_0^t{\gamma }^{-\alpha }}{B}_{n,m}\left(t-\gamma \right)e^{C_{n,m}^{(2)}\left(\gamma -t\right)}d\gamma -C_{n,m}^{(2)}{\displaystyle {\int }_0^t{\gamma }^{-\alpha }}{\displaystyle {\int }_0^{t-\gamma }{B}_{n,m}\left(\tau \right)d\tau }{e}^{C_{n,m}^{(2)}\left(\gamma -t\right)}d\gamma \right)\\ -B_{n,m}\left(t\right)e^{-C_{n,m}^{(2)}t}=C_{n,m}^{(3)}g\left(t\right).\end{array}$$

(29)

Denote $t=0$, we have the relationship:

$$B_{n,m}\left(0\right)=-C_{n,m}^{(3)}g\left(0\right).$$

(30)

Multiplying the left and right sides of (29) by $e^{C_{n,m}^{(2)}t}$ and taking the derivative of $t$, yields:

$$\begin{array}{l}\frac{C_{n,m}^{(1)}}{\mathsf{\Gamma }\left(1-\alpha \right)}\left(t^{-\alpha }{B}_{n,m}\left(0\right)e^{C_{n,m}^{(2)}t}+{\displaystyle {\int }_0^t{\gamma }^{-\alpha }}{B}_{n,m}{}^{\prime }\left(t-\gamma \right)e^{C_{n,m}^{(2)}\gamma }d\gamma -C_{n,m}^{(2)}{\displaystyle {\int }_0^t{\gamma }^{-\alpha }}{B}_{n,m}\left(t-\gamma \right)e^{C_{n,m}^{(2)}\gamma }d\gamma \right)\\ -\frac{d{B}_{n,m}\left(t\right)}{dt}=e^{C_{n,m}^{(2)}t}{C}_{n,m}^{(3)}\left(C_{n,m}^{(2)}g\left(t\right)+g^{\prime }\left(t\right)\right).\end{array}$$

(31)

Resort to variable $\gamma =t-\xi$ and multiply both sides of Equation (31) by $e^{-C_{n,m}^{(2)}t}$, we have:

$$\begin{array}{l}\frac{C_{n,m}^{(1)}}{\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle {\int }_0^t{\left(t-\xi \right)}^{-\alpha }}\frac{d{B}_{n,m}\left(\xi \right)e^{-C_{n,m}^{(2)}\xi }}{d\xi }d\xi -\frac{d{B}_{n,m}\left(t\right)e^{-C_{n,m}^{(2)}t}}{dt}-C_{n,m}^{(2)}{B}_{n,m}\left(t\right)e^{-C_{n,m}^{(2)}t}\\ =C_{n,m}^{(3)}\left(C_{n,m}^{(2)}g\left(t\right)+g^{\prime }\left(t\right)\right)-\frac{C_{n,m}^{(1)}}{\mathsf{\Gamma }\left(1-\alpha \right)}{t}^{-\alpha }{B}_{n,m}\left(0\right).\end{array}$$

(32)

Denoting $A_{n,m}\left(t\right)=B_{n,m}\left(t\right)e^{-C_{n,m}^{(2)}t}$ and according to (30), we have

$$C_{n,m}^{(1)}\frac{d^{\alpha +1}{A}_{n,m}}{d{t}^{\alpha +1}}-\frac{d{A}_{n,m}}{dt}-C_{n,m}^{(2)}{A}_{n,m}=C_{n,m}^{(3)}\left(C_{n,m}^{(2)}g\left(t\right)+g^{\prime }\left(t\right)\right)-\frac{C_{n,m}^{(1)}{A}_{n,m}\left(0\right)}{\mathsf{\Gamma }\left(1-\alpha \right)}{t}^{-\alpha },$$

(33)

where $n=2k_1-1$, $m=2k_2-1$ and $A_{n,m}\left(0\right)=-C_{n,m}^{(3)}g\left(0\right)$. The analytical solution can be obtained by referring to [46].

Then the solution to Equation (20) can be obtained as:

$$w\left(x,y,t\right)={\displaystyle \sum _{\begin{array}{l}\hspace{1em}n=1\\ n=2k_1-1\end{array}}^{+\infty }{\displaystyle \sum _{\begin{array}{l}\hspace{1em}m=1\\ m=2k_2-1\end{array}}^{+\infty }{\displaystyle {\int }_0^t{B}_{n,m}\left(\tau \right)d\tau }\cdot e^{-\left(v{\pi }^2\frac{n^2+m^2}{4}+M\right)t}\sin \left(\frac{n\pi }{2}\left(x+1\right)\right)\sin \left(\frac{m\pi }{2}\left(y+1\right)\right)}},$$

(34)

where $B_{n,m}\left(t\right)=e^{C_{n,m}^{(2)}t}{A}_{n,m}\left(t\right)$ and $A_{n,m}\left(t\right)$ refers to the solution of (33).

4. Numerical Discretization Method

Numerical Scheme

Firstly, we divide the spatial region $\left[-a,a\right]\times \left[-b,b\right]$ with the uniform mesh points $x_i=-a+i{h}_x$, $i=0,1,\cdots ,M_x$, $y_j=-b+j{h}_y$, $j=0,1,\cdots ,M_y$, in which $h_x=2a/M_x$, $h_y=2b/M_y$. For the time region $\left[0,T\right]$, we take $t_n=n\tau$ with time step $\tau =T/N$ for $n=0,1,\cdots N$. Define ${\mathsf{\Omega }}_h\equiv \left\{\left(x_i,y_j\right)|0\le i\le M_x,0\le j\le M_y\right\}$ and ${\mathsf{\Omega }}_{\tau }\equiv \left\{t_n|0\le n\le N\right\}$.

For a net function $w=\left\{w_{i,j}^n|0\le i\le M_x,0\le j\le M_y,0\le n\le N\right\}$ defined on an interval ${\mathsf{\Omega }}_h\times {\mathsf{\Omega }}_{\tau }$, denote the following symbols for simplicity:

$${\nabla }_t{w}_{i,j}^n=\frac{w_{i,j}^n-w_{i,j}^{n-1}}{\tau }, {\delta }_x{w}_{i,j}^n=\frac{w_{i,j}^n-w_{i-1,j}^n}{h_x}, {\delta }_y{w}_{i,j}^n=\frac{w_{i,j}^n-w_{i,j-1}^n}{h_y},$$

$${\delta }_x^2{w}_{i,j}^n=\frac{w_{i+1,j}^n-2w_{i,j}^n+w_{i-1,j}^n}{h_x^2}, {\delta }_y^2{w}_{i,j}^n=\frac{w_{i,j+1}^n-2w_{i,j}^n+w_{i,j-1}^n}{h_x^2}$$

Furthermore, the exact solution is defined as $W_{i,j}^n=w\left(x_i,y_j,t_n\right)$ for simplicity. Applying the L1-scheme [37] for discretizing the fractional derivative, at the mesh points $\left(x_i,y_j,t_n\right)$, we have:

$$\frac{{\partial }^{\alpha }{W}_{i,j}^n}{\partial t^{\alpha }}=\frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\left(c_0{W}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{W}_{i,j}^k-c_{n-1}{W}_{i,j}^0\right)+{\left(R_1\right)}_{i,j}^n$$

(35)

where $c_k={\left(k+1\right)}^{1-\alpha }-k^{1-\alpha }$ and $\left|{\left(R_1\right)}_{i,j}^n\right|\le C{\tau }^{2-\alpha }$.

At the mesh points $\left(x_i,y_j,t_n\right)$, the backward difference method is applied to discretize the time derivative of order one

$$\frac{\partial W_{i,j}^n}{\partial t}={\nabla }_t{W}_{i,j}^n+\mathrm{O}\left(\tau \right)$$

(36)

Use of the central difference scheme yields the discretization schemes for the second order space derivatives

$$\frac{{\partial }^2{W}_{i,j}^n}{\partial x^2}={\delta }_x^2{W}_{i,j}^n+\mathrm{O}\left(h_x^2\right) \mathrm{and} \frac{{\partial }^2{W}_{i,j}^n}{\partial y^2}={\delta }_y^2{W}_{i,j}^n+\mathrm{O}\left(h_y^2\right)$$

(37)

Combining (35) and (37), we have the difference schemes for the mixed derivatives of time and space:

$$\frac{{\partial }^{\alpha }}{\partial t^{\alpha }}\frac{{\partial }^2}{\partial x^2}{W}_{i,j}^n=\frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\left(c_0{\delta }_x^2{W}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_x^2{W}_{i,j}^k-c_{n-1}{\delta }_x^2{W}_{i,j}^0\right)+{\left(R_2\right)}_{i,j}^n,$$

(38)

$$\frac{{\partial }^{\alpha }}{\partial t^{\alpha }}\frac{{\partial }^2}{\partial y^2}{W}_{i,j}^n=\frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\left(c_0{\delta }_y^2{W}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_y^2{W}_{i,j}^k-c_{n-1}{\delta }_y^2{W}_{i,j}^0\right)+{\left(R_3\right)}_{i,j}^n,$$

(39)

where $\left|{\left(R_2\right)}_{i,j}^n\right|\le C\left({\tau }^{2-\alpha }+h_x^2\right)$ and $\left|{\left(R_3\right)}_{i,j}^n\right|\le C\left({\tau }^{2-\alpha }+h_y^2\right)$.

Denote the discretization scheme for $-\frac{1}{\rho }\frac{\partial p}{\partial z}$ at the points $\left(x_i,y_j,t_n\right)$ as $g_{i,j}^n$. Through the difference schemes (35)–(39), we have the final discretization scheme for the governing Equation (8)

$$\begin{array}{l}{\nabla }_t{W}_{i,j}^n+M{W}_{i.j}^n-\nu {\delta }_x^2{W}_{i,j}^n-\nu {\delta }_y^2{W}_{i,j}^n\\ =\nu \lambda \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\left(c_0{\delta }_x^2{W}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_x^2{W}_{i,j}^k-c_{n-1}{\delta }_x^2{W}_{i,j}^0\right)\\ +\nu \lambda \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\left(c_0{\delta }_y^2{W}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_y^2{W}_{i,j}^k-c_{n-1}{\delta }_y^2{W}_{i,j}^0\right)+g_{i,j}^n+R_{i,j}^n,\end{array}$$

(40)

where $\left|R_{i,j}^n\right|\le C\left(\tau +h_x^2+h_y^2\right)$.

Substituting $W_{i,j}^n$ with $w_{i,j}^n$, we have the numerical difference scheme of Equation (8)

$$\begin{array}{l}{\nabla }_t{w}_{i,j}^n+M{w}_{i.j}^n-\nu {\delta }_x^2{w}_{i,j}^n-\nu {\delta }_y^2{w}_{i,j}^n\\ =\nu \lambda \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\left(c_0{\delta }_x^2{w}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_x^2{w}_{i,j}^k-c_{n-1}{\delta }_x^2{w}_{i,j}^0\right)\\ +\nu \lambda \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\left(c_0{\delta }_y^2{w}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_y^2{w}_{i,j}^k-c_{n-1}{\delta }_y^2{w}_{i,j}^0\right)+g_{i,j}^n.\end{array}$$

(41)

By merging the terms at the same time layer, making the left side the $n$-th time layer, and the right side the time layer with the order less than $n$, Equation (41) can be rewritten in another form:

$$\begin{array}{l}\left(\frac{1}{\tau }+M\right)w_{i,j}^n-\frac{\nu }{h_x^2}\left(r_1+1\right)\left(w_{i+1,j}^n-2w_{i,j}^n+w_{i-1,j}^n\right)-\frac{\nu }{h_y^2}\left(r_1+1\right)\left(w_{i,j+1}^n-2w_{i,j}^n+w_{i.j-1}^n\right)\\ =\frac{1}{\tau }{w}_{i,j}^{n-1}-\nu r_1\left[{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}\left({\delta }_x^2{w}_{i,j}^k+{\delta }_y^2{w}_{i,j}^k\right)+c_{n-1}\left({\delta }_x^2{w}_{i,j}^0+{\delta }_y^2{w}_{i,j}^0\right)\right]+g_{i,j}^n,\end{array}$$

(42)

where $r_1=\frac{\lambda {\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}$ and $g_{i,j}^n=-\frac{1}{\rho }\frac{\partial p\left(x_i,y_j,t_n\right)}{\partial z}$.

In what follows, the symbol $E$ denotes the unit matrix which may be with a different order in various sections. Considering the zero-boundary conditions, the discretization scheme (42) can be rewritten in a matrix form:

$$\begin{array}{l}\left(\frac{1}{\tau }+M\right)E{w}^n-\frac{\nu \left(r_1+1\right)}{h_x^2}E\otimes K_1{w}^n-\frac{\nu \left(r_1+1\right)}{h_y^2}{K}_2\otimes E{w}^n\\ =\frac{1}{\tau }E{w}^{n-1}-\nu r_1\left[{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}\left(E\otimes K_1+K_2\otimes E\right)w^k+c_{n-1}\left(E\otimes K_1+K_2\otimes E\right)w^0\right]+g^n,\end{array}$$

where the symbol $\otimes$ denotes the Kronecker product [47],

$$K_1={\left(\begin{array}{ccccc}-2& 1& & & \\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\end{array}\right)}_{\left(M_x-1\right)\times \left(M_x-1\right)}, K_2={\left(\begin{array}{ccccc}-2& 1& & & \\ 1& -2& 1& & \\ & \ddots & \ddots & \ddots & \\ & & 1& -2& 1\end{array}\right)}_{\left(M_y-1\right)\times \left(M_y-1\right)},$$

$$w^n={\left(w_{1,1}^n,w_{2,1}^n,\dots ,w_{M_x-1,1}^n,w_{1,2}^n,w_{2,2}^n,\dots ,w_{M_x-1,2}^n,\dots ,w_{1,M_y-1}^n,w_{2,M_y-1}^n,\dots ,w_{M_x-1,M_y-1}^n\right)}^T,$$

$$g^n={\left(g_{1,1}^n,g_{2,1}^n,\dots ,g_{M_x-1,1}^n,g_{1,2}^n,g_{2,2}^n,\dots ,g_{M_x-1,2}^n,\dots ,g_{1,M_y-1}^n,g_{2,M_y-1}^n,\dots ,g_{M_x-1,M_y-1}^n\right)}^T.$$

The initial condition can be discretized as $w_{i,j}^0=0$ and the boundary conditions are discretized as $w_{0,j}^n=w_{M_x,j}^n=w_{i,0}^n=w_{i,M_y}^n=0$. The above numerical method can be applied to widespread situations, for example, the dynamics in porous media for solving Richards’ equation [48]. For this equation, the treating method mentioned above can be similarly applied.

Besides the velocity distribution, the shear force is another important quantity to analyze. We consider the shear force ${\tau }_{xz}$ for $xz$-direction at the wall surface ($x=0$), and the difference scheme is given as:

$$\begin{array}{l}{\tau }_{xz}=\left(\mu +{\alpha }_1{D}_t^{\alpha }\right){\frac{\partial w}{\partial x}|}_{x=0}\\ \approx \left[\mu +\frac{{\alpha }_1{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\right]\frac{w_{1,j}^n-w_{0,j}^n}{h_x}-\frac{{\alpha }_1{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}\frac{w_{1,j}^k-w_{0,j}^k}{h_x}-\frac{{\alpha }_1{\tau }^{-\alpha }{c}_{n-1}}{\mathsf{\Gamma }\left(2-\alpha \right)}\frac{w_{1,j}^0-w_{0,j}^0}{h_x}.\end{array}$$

Due to the symmetry of the velocity in the $x$- and $y$-directions, we deduce the shear force along the $yz$-direction at the wall surface $y=0$ to be the same as the $xz$-direction.

5. Feasibility Analysis

Denote $V_h=\left\{v|v\right\}$ is a net function on ${\mathsf{\Omega }}_h\times {\mathsf{\Omega }}_{\tau }$, $v_{i,j}^n=0$ when $i=0$ and $M_x$ or $j=0$ and $M_y$. For $w^n$, $v^n\in V_h$, we denote the discrete inner products and norms:

$$\left(w^n,v^n\right)=h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{w}_{i,j}^n{v}_{i,j}^n}} \mathrm{and} {\parallel w^n\parallel }^2=\left(w^n,w^n\right).$$

(43)

Lemma 1.

[49] The matrix$A\otimes B$is symmetric positive definite with the condition that both$A\in {ℝ}^{n\times n}$and$B\in {ℝ}^{n\times n}$satisfy symmetric positive definite. For$\forall 0\ne v\in {ℝ}^{n^2}$, it holds that:

$$v^T\left(A\otimes B\right)v>0.$$

(44)

Lemma 2.

[50] For all $A$and$B$, ${\left(A\otimes B\right)}^T=A^T\otimes B^T$.

Lemma 3.

For $w,v\in {\mathsf{\Omega }}_h\times {\mathsf{\Omega }}_{\tau }$, it is straightforward to check that$\left({\delta }_x^2{w}^k,v^k\right)=-\left({\delta }_x{w}^k,{\delta }_x{v}^k\right)$with the zero-boundary conditions by applying integration by parts.

Lemma 4.

[37] For the symbols $c_j$in (35), define the vector$S={\left[S_1,S_2,\dots ,S_N\right]}^T$and constant$P$, it holds that:

$$\frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}{\displaystyle \sum _{k=1}^N\left[c_0{S}_k-{\displaystyle \sum _{j=1}^{k-1}\left(c_{k-j-1}-c_{k-j}\right)S_j}-c_{k-1}P\right]S_k}\ge \frac{T^{-\alpha }}{2\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle \sum _{k=1}^N{S}_k^2}-\frac{T^{1-\alpha }}{2\tau \mathsf{\Gamma }\left(2-\alpha \right)}{P}^2$$

5.1. Solvability

Theorem 2.

Denote $w_{i,j}^n$as the numerical solution of Equations (8)–(10) for$i=0,1,\cdots ,M_x$, $j=0,1,\cdots ,M_y$and$n=0,1,\cdots N$, then (42) is uniquely solvable.

Proof. 

Denote the coefficient matrix $G=\left(\frac{1}{\tau }+M\right)E-\frac{\nu \left(r_1+1\right)}{h_x^2}E\otimes K_1-\frac{\nu \left(r_1+1\right)}{h_y^2}{K}_2\otimes E$.

Firstly, using Lemma 3, we have:

$$G^T=\left(\frac{1}{\tau }+M\right)E^T-\frac{\nu \left(r_1+1\right)}{h_x^2}{E}^T\otimes K_1^T-\frac{\nu \left(r_1+1\right)}{h_y^2}{K}_2^T\otimes E^T=G$$

Furthermore, the matrix $G$ can simply be verified as strictly diagonally dominant. Then, the matrix $G$ is positive definite. Therefore, the numerical difference scheme has a unique solution. □

5.2. Stability

Theorem 3.

The scheme (41) possesses unconditional stability, which satisfies:

$${\Vert w_{i,j}^N\Vert }^2\le \frac{T}{2M}\underset{1\le n\le N}{\max }{\Vert g_{i,j}^n\Vert }^2$$

Proof. 

Multiplying both sides of Equation (41) by $\tau h_x{h}_y{w}_{i,j}^n$, and summing $i$, $j$, $n$ from $1$ to $M_x-1$, $1$ to $M_y-1$, $1$ to $N$, respectively, we derive the following equation:

$$\begin{array}{l}\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{w}_{i,j}^n}}}{\nabla }_t{w}_{i,j}^n+M\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{w}_{i,j}^n}}}{w}_{i.j}^n-\nu \tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{w}_{i,j}^n}}}\left({\delta }_x^2{w}_{i,j}^n-{\delta }_y^2{w}_{i,j}^n\right)\\ -\frac{\nu \lambda {\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N\left(c_0{\delta }_x^2{w}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_x^2{w}_{i,j}^k-c_{n-1}{\delta }_x^2{w}_{i,j}^0\right)w_{i,j}^n}}}\\ -\frac{\nu \lambda {\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N\left(c_0{\delta }_y^2{w}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_y^2{w}_{i,j}^k-c_{n-1}{\delta }_y^2{w}_{i,j}^0\right)w_{i,j}^n}}}\\ =\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{w}_{i,j}^n}}}{g}_{i,j}^n.\end{array}$$

By applying the inequation $a\left(a-b\right)\ge \frac{1}{2}\left(a^2-b^2\right)$ and considering the zero initial condition, the first term satisfies:

$$\begin{array}{l}\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{w}_{i,j}^n}}}{\nabla }_t{w}_{i,j}^n\ge \frac{1}{2}{h}_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N\left[{\left(w_{i,j}^n\right)}^2-{\left(w_{i,j}^{n-1}\right)}^2\right]}}}\\ =\frac{1}{2}{h}_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}\left[{\left(w_{i,j}^N\right)}^2-{\left(w_{i,j}^0\right)}^2\right]}}=\frac{1}{2}\left({\Vert w^N\Vert }^2-{\Vert w^0\Vert }^2\right)=\frac{1}{2}{\Vert w^N\Vert }^2.\end{array}$$

Considering the relationship between the norm and inner product, the second term yields

$$M\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{w}_{i,j}^n}}}{w}_{i.j}^n=M\tau {\displaystyle \sum _{n=1}^N\left(w_{i,j}^n,w_{i,j}^n\right)}=M\tau {\displaystyle \sum _{n=1}^N{\Vert w^n\Vert }^2}$$

By using the Lemma 3, for the third term, we have

$$\begin{array}{l}-\nu \tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{w}_{i,j}^n}}}\left({\delta }_x^2{w}_{i,j}^n+{\delta }_y^2{w}_{i,j}^n\right)\\ =\nu \tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{\delta }_x{w}_{i,j}^n}}}{\delta }_x{w}_{i,j}^n+\nu \tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{\delta }_y{w}_{i,j}^n}}}{\delta }_y{w}_{i,j}^n\\ =\nu \tau {\displaystyle \sum _{n=1}^N{\Vert {\delta }_x{w}^n\Vert }^2}+\nu \tau {\Vert {\delta }_y{w}^n\Vert }^2\ge 0\end{array}$$

.

By applying Lemma 4, the fourth term satisfies:

$$\begin{array}{l}-\frac{\nu \lambda {\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N\left(c_0{\delta }_x^2{w}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_x^2{w}_{i,j}^k-c_{n-1}{\delta }_x^2{w}_{i,j}^0\right)w_{i,j}^n}}}\\ =\frac{\nu \lambda {\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N\left(c_0{\delta }_x{w}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_x{w}_{i,j}^k-c_{n-1}{\delta }_x{w}_{i,j}^0\right){\delta }_x{w}_{i,j}^n}}}\\ \ge \nu \lambda \tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}\left[\frac{T^{-\alpha }}{2\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle \sum _{n=1}^N{\left({\delta }_x{w}_{i,j}^n\right)}^2}-\frac{T^{1-\alpha }}{2\tau \mathsf{\Gamma }\left(2-\alpha \right)}{\left({\delta }_x{w}_{i,j}^0\right)}^2\right]}}\\ =\frac{\nu \lambda \tau T^{-\alpha }}{2\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle \sum _{n=1}^N{\Vert {\delta }_x{w}^n\Vert }^2}-\frac{\nu \lambda T^{1-\alpha }}{2\mathsf{\Gamma }\left(2-\alpha \right)}{\Vert {\delta }_x{w}^0\Vert }^2\ge 0.\end{array}$$

Similarly, for the fifth term, it satisfies

$$-\frac{\nu \lambda {\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N\left(c_0{\delta }_y^2{w}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_y^2{w}_{i,j}^k-c_{n-1}{\delta }_y^2{w}_{i,j}^0\right)w_{i,j}^n}}}\ge 0.$$

By using the Cauchy–Schwartz inequality, the last term changes as:

$$\tau h_x{h}_y{\displaystyle \sum _{i=1}^{M_x-1}{\displaystyle \sum _{i=1}^{M_y-1}{\displaystyle \sum _{n=1}^N{w}_{i,j}^n}}}{g}_{i,j}^n=\tau {\displaystyle \sum _{n=1}^N\left(w_{i,j}^n,g_{i,j}^n\right)}\le M\tau {\displaystyle \sum _{n=1}^N{\Vert w^n\Vert }^2}+\frac{\tau }{4M}{\displaystyle \sum _{n=1}^N{\Vert g^n\Vert }^2}.$$

As a conclusion, we deduce:

$${\Vert w^N\Vert }^2\le \frac{\tau }{2M}{\displaystyle \sum _{n=1}^N{\Vert g^n\Vert }^2}\le \frac{T}{2M}\underset{1\le n\le N}{\max }{\Vert g^n\Vert }^2.$$

□

5.3. Convergence

Define the error $e_{i,j}^n=w_{i,j}^n-w\left(x_i,y_j,t_n\right)$. Taking the difference between the Equations (40) and (41), we deduce that the error satisfies:

$$\begin{array}{l}{\nabla }_t{e}_{i,j}^n+M{e}_{i.j}^n-\nu {\delta }_x^2{e}_{i,j}^n-\nu {\delta }_y^2{e}_{i,j}^n\\ =\nu \lambda \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\left(c_0{\delta }_x^2{e}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_x^2{e}_{i,j}^k-c_{n-1}{\delta }_x^2{e}_{i,j}^0\right)\\ +\nu \lambda \frac{{\tau }^{-\alpha }}{\mathsf{\Gamma }\left(2-\alpha \right)}\left(c_0{\delta }_y^2{e}_{i,j}^n-{\displaystyle \sum _{k=1}^{n-1}\left(c_{n-k-1}-c_{n-k}\right)}{\delta }_y^2{e}_{i,j}^k-c_{n-1}{\delta }_y^2{e}_{i,j}^0\right)+\mathrm{O}\left(\tau +h_x^2+h_y^2\right).\end{array}$$

(45)

Theorem 4.

The scheme (41) is convergent with the following form:

$${\Vert e^N\Vert }^2\le \frac{T}{2M}{\left(\tau +h_x^2+h_y^2\right)}^2.$$

(46)

Proof. 

Similar to the proof of the stability, substituting the source term with the error, we have:

$${\Vert e^N\Vert }^2\le \frac{\tau }{2M}{\displaystyle \sum _{n=1}^N{\left(\tau +h_x^2+h_y^2\right)}^2}=\frac{T}{2M}{\left(\tau +h_x^2+h_y^2\right)}^2.$$

(47)

□

6. Acceleration of the Fractional Derivative

The traditional treating method for the fractional derivative is to use the L1 scheme with an expensive cost of computation and storage due to the non-locality that the fractional derivative contains. The difference scheme at $t=t_n$ contains a summation of all values from zero to the current time and the total cost at every spatial point is $\mathrm{O}\left(N^2\right)$. To reduce the computational and storage cost, a fast algorithm [38] is applied. Here we summarized the main idea of the fast algorithm.

The definition of Caputo’s fractional derivative of order $0<\alpha <1$ can be expressed as the summation of two terms, a local part $C_l\left(t_n\right)$ and a history part $C_h\left(t_n\right)$:

$$\begin{array}{cc}\hfill {{}^{C}D_t^{\alpha }w\left(t\right)|}_{t=t_n}& =\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle {\int }_0^{t_n}{\left(t_n-s\right)}^{-\alpha }\frac{\partial w\left(s\right)}{\partial s}}ds\hfill \\ & =\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle {\int }_{t_{n-1}}^{t_n}\frac{1}{{\left(t_n-s\right)}^{\alpha }}\frac{\partial w\left(s\right)}{\partial s}}ds+\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle {\int }_0^{t_{n-1}}\frac{1}{{\left(t_n-s\right)}^{\alpha }}}\frac{\partial w\left(s\right)}{\partial s}ds\hfill \\ & :=C_l\left(t_n\right)+C_h\left(t_n\right).\hfill \end{array}$$

(48)

For the local portion, we approximate $\frac{\partial w\left(s\right)}{\partial s}$ by $\frac{w\left(t_n\right)-w\left(t_{n-1}\right)}{\tau }$, yields

$$C_l(t_n)\approx \frac{w\left(t_n\right)-w\left(t_{n-1}\right)}{\tau \mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle {\int }_{t_{n-1}}^{t_n}\frac{ds}{{\left(t_n-s\right)}^{\alpha }}}=\frac{w\left(t_n\right)-w\left(t_{n-1}\right)}{{\tau }^{\alpha }\mathsf{\Gamma }\left(2-\alpha \right)}.$$

(49)

We employ the integration by parts for the history part

$$C_h(t_n)=\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}\left[\frac{w\left(t_{n-1}\right)}{{\tau }^{\alpha }}-\frac{w\left(t_0\right)}{t_n^{\alpha }}-\alpha {\displaystyle {\int }_0^{t_{n-1}}\frac{w\left(s\right)}{{\left(t_n-s\right)}^{\alpha +1}}ds}\right].$$

(50)

Treating the kernel $\frac{1}{t^{\alpha +1}}$ in the convolution integral is the key. Referring to [38], for any time interval $\left[\tau ,T\right]$, the kernel $\frac{1}{t^{\alpha +1}}$ can be approached by an efficient sum-of-exponentials approximation with a prescribed absolute error $\epsilon$. Specifically speaking, there are real positive numbers $w_l$ and $s_l$ ($l=1,\cdots ,N_{\exp }$) such that

$$\left|\frac{1}{t^{\alpha +1}}-{\displaystyle \sum _{l=1}^{N_{\exp }}{\omega }_l{e}^{-s_{l}t}}\right|\le \epsilon , \mathrm{for} \mathrm{any} t\in \left[\tau ,T\right],$$

(51)

where $N_{\exp }$ is of the order

$$N_{\exp }=\mathcal{O}\left(\log \frac{1}{\epsilon }\left(\log \log \frac{1}{\epsilon }+\log \frac{T}{\sigma }\right)+\log \frac{1}{\sigma }\left(\log \log \frac{1}{\epsilon }+\log \frac{1}{\sigma }\right)\right).$$

(52)

Equation (4) is the main idea for the fast algorithm. The sum-of-exponentials approximation for the kernel $\frac{1}{t^{\beta }}$ can also be generalized for the order $0<\alpha <2$ [38,51].

We substitute the kernel $\frac{1}{t^{\alpha +1}}$ via the formular (51) to approximate the history portion as:

$$C_h\left(t_n\right)\approx \frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}\left[\frac{w\left(t_{n-1}\right)}{{\tau }^{\alpha }}-\frac{w\left(t_0\right)}{t_n^{\alpha }}-\alpha {\displaystyle \sum _{l=1}^{N_{\exp }}{w}_l}{W}_{his,l}(t_n)\right],$$

(53)

where $W_{his,l}\left(t_n\right)={\displaystyle {\int }_0^{t_{n-1}}{e}^{-(t_n-t)s_l}w\left(t\right)dt}$.

The function $W_{his,l}\left(t_n\right)$ is calculated for $n=1,2,\cdots ,N$ and the following recurrent relationship is derived

$$W_{hist,l}\left(t_n\right)=e^{-s_l\tau }{W}_{hist,l}\left(t_{n-1}\right)+{\displaystyle {\int }_{t_{n-2}}^{t_{n-1}}{e}^{-s_l\left(t_n-\tau \right)}w\left(\tau \right)d\tau }, W_{hist,l}\left(t_0\right)=0.$$

(54)

The integral in (54) could be rewritten as:

$${\displaystyle {\int }_{t_{n-2}}^{t_{n-1}}{e}^{-s_l\left(t_n-\tau \right)}w\left(\tau \right)d\tau }\approx \frac{e^{-s_l\tau }}{s_l^2\tau }\left[\left(e^{-s_l\tau }-1+s_l\tau \right)w^{n-1}+\left(1-e^{-s_l\tau }-e^{-s_l\tau }{s}_l\tau \right)w^{n-2}\right].$$

(55)

To compute $W_{his,i}\left(t_n\right)$, as Equation (55) indicates, $W_{his,i}\left(t_{n-1}\right)$ is already computed and stored and the cost is needed by only $\mathrm{O}\left(1\right)$ at each step. As (6.4) indicates, the cost to evaluate the fractional derivative is needed $\mathrm{O}\left(N_{\exp }\right)$ at each time step. That is to say, a reduction from $\mathrm{O}\left(N\right)$ to $\mathrm{O}\left(\log N\right)$ or $\mathrm{O}\left({\log }^{2}N\right)$.

As a summation, the fast evolution of the Caputo’s fractional derivative at $t=t_n$ is given as:

$${}^{F}D_t^{\alpha }w\left(x,y,t_n\right)=\frac{W^n-W^{n-1}}{{\tau }^{\alpha }\mathsf{\Gamma }\left(2-\alpha \right)}+\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}\left[\frac{W^{n-1}}{{\tau }^{\alpha }}-\frac{W^0}{t_n^{\alpha }}-\alpha {\displaystyle \sum _{l=1}^{N_{\exp }}{\omega }_l{W}_{hist,l}\left(t_n\right)}\right]+R_1,$$

(56)

where $\left|R_1\right|\le C\left({\tau }^{2-\alpha }+\epsilon \right)$ and the recurrence relation satisfies (6.7) and (6.8).

Combining (56) and (37), we have:

$$\frac{{\partial }^{\alpha }}{\partial t^{\alpha }}\frac{{\partial }^2}{\partial x^2}{w}_{i,j}^n=\frac{{\delta }_x^2{w}_{i,j}^n-{\delta }_x^2{w}_{i,j}^{n-1}}{{\tau }^{\alpha }\mathsf{\Gamma }\left(2-\alpha \right)}+\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}\left[\frac{{\delta }_x^2{w}_{i,j}^{n-1}}{{\tau }^{\alpha }}-\frac{{\delta }_x^2{w}_{i,j}^0}{t_n^{\alpha }}-\alpha {\displaystyle \sum _{l=1}^{N_{\exp }}{\omega }_l{\delta }_x^2{w}_{hist,l}\left(t_n\right)}\right],$$

(57)

$$\frac{{\partial }^{\alpha }}{\partial t^{\alpha }}\frac{{\partial }^2}{\partial y^2}{w}_{i,j}^n=\frac{{\delta }_y^2{w}_{i,j}^n-{\delta }_y^2{w}_{i,j}^{n-1}}{{\tau }^{\alpha }\mathsf{\Gamma }\left(2-\alpha \right)}+\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}\left[\frac{{\delta }_y^2{w}_{i,j}^{n-1}}{{\tau }^{\alpha }}-\frac{{\delta }_y^2{w}_{i,j}^0}{t_n^{\alpha }}-\alpha {\displaystyle \sum _{l=1}^{N_{\exp }}{\omega }_l{\delta }_y^2{w}_{hist,l}\left(t_n\right)}\right],$$

(58)

where ${\delta }_x^2{w}_{hist,l}\left(t_n\right)=e^{-s_l\tau }{\delta }_x^2{w}_{hist,l}\left(t_{n-1}\right)+{\displaystyle {\int }_{t_{n-2}}^{t_{n-1}}{e}^{-s_l\left(t_n-\tau \right)}{\delta }_x^{2}w\left(\tau \right)d\tau }$, ${\delta }_x^2{w}_{hist,l}\left(t_0\right)=0$, ${\displaystyle {\int }_{t_{n-2}}^{t_{n-1}}{e}^{-s_l\left(t_n-\tau \right)}{\delta }_x^{2}w\left(\tau \right)d\tau }\approx \frac{e^{-s_i\tau }}{s_i^2\tau }\left[\left(e^{-s_i\tau }-1+s_i\tau \right){\delta }_x^2{w}^{n-1}+\left(1-e^{-s_i\tau }-e^{-s_i\tau }{s}_i\tau \right){\delta }_x^2{w}^{n-2}\right]$, ${\delta }_y^2{w}_{hist,l}\left(t_n\right)=e^{-s_l\tau }{\delta }_y^2{w}_{hist,l}\left(t_{n-1}\right)+{\displaystyle {\int }_{t_{n-2}}^{t_{n-1}}{e}^{-s_l\left(t_n-\tau \right)}{\delta }_y^{2}w\left(\tau \right)d\tau }$, ${\delta }_y^2{w}_{hist,l}\left(t_0\right)=0$,${\displaystyle {\int }_{t_{n-2}}^{t_{n-1}}{e}^{-s_l\left(t_n-\tau \right)}{\delta }_y^{2}w\left(\tau \right)d\tau }\approx \frac{e^{-s_i\tau }}{s_i^2\tau }\left[\left(e^{-s_i\tau }-1+s_i\tau \right){\delta }_y^2{w}^{n-1}+\left(1-e^{-s_i\tau }-e^{-s_i\tau }{s}_i\tau \right){\delta }_y^2{w}^{n-2}\right]$.

By a combination, we deduce the final difference scheme:

$$\begin{array}{l}\left(\frac{1}{\tau }+M\right)w_{i,j}^n-\frac{\nu r_2}{h_x^2}\left(w_{i+1,j}^n-2w_{i,j}^n+w_{i-1,j}^n\right)-\frac{\nu r_2}{h_y^2}\left(w_{i,j+1}^n-2w_{i,j}^n+w_{i,j-1}^n\right)\\ =\frac{1}{\tau }{w}_{i,j}^{n-1}+\frac{\nu \lambda }{{\tau }^{\alpha }}\left(\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}-\frac{1}{\mathsf{\Gamma }\left(2-\alpha \right)}\right)\left({\delta }_x^2{w}_{i,j}^{n-1}+{\delta }_y^2{w}_{i,j}^{n-1}\right)\\ -\frac{\nu \lambda }{\mathsf{\Gamma }\left(1-\alpha \right)t_n^{\alpha }}\left({\delta }_x^2{w}_{i,j}^0+{\delta }_y^2{w}_{i,j}^0\right)-\frac{\nu \lambda \alpha }{\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle \sum _{l=1}^{N_{\exp }}{\omega }_l\left[{\delta }_x^2{w}_{hist,l}\left(t_n\right)+{\delta }_y^2{w}_{hist,l}\left(t_n\right)\right]}+g_{i,j}^n,\end{array}$$

(59)

where $r_2=\frac{\lambda }{{\tau }^{\alpha }\mathsf{\Gamma }\left(2-\alpha \right)}+1$.

The discretization scheme (59) can be rewritten in a matrix form:

$$\begin{array}{l}\left[\left(\frac{1}{\tau }+M\right)E-\frac{\nu r_2}{h_x^2}E\otimes K_1-\frac{\nu r_2}{h_y^2}{K}_2\otimes E\right]w^n\\ =\left[\frac{1}{\tau }E+\frac{\nu \lambda }{{\tau }^{\alpha }}\left(\frac{1}{\mathsf{\Gamma }\left(1-\alpha \right)}-\frac{1}{\mathsf{\Gamma }\left(2-\alpha \right)}\right)\left(E\otimes K_1+K_2\otimes E\right)\right]w^{n-1}+g^n\\ -\frac{\nu \lambda }{\mathsf{\Gamma }\left(1-\alpha \right)t_n^{\alpha }}\left(E\otimes K_1+K_2\otimes E\right)w^0-\frac{\alpha \nu \lambda }{\mathsf{\Gamma }\left(1-\alpha \right)}{\displaystyle \sum _{l=1}^{N_{\exp }}{\omega }_l\left[{\delta }_x^2{w}_{hist,l}\left(t_n\right)+{\delta }_y^2{w}_{hist,l}\left(t_n\right)\right]}.\end{array}$$

(60)

7. Results and Discussion

Example 1.

(Verification of the discretization scheme).

The governing equation is solved numerically that the fractional derivative is discretized by the traditional L1 difference method and the fast algorithm. How to verify the correctness of the difference method is the key. As Section 3 indicates, the exact solution is complicated. As a modification, a source term is introduced and the governing equation changes as:

$$\frac{\partial w}{\partial t}=\nu \left(1+\lambda \frac{D^{\alpha }}{D{t}^{\alpha }}\right)\left(\frac{{\partial }^{2}w}{\partial x^2}+\frac{{\partial }^{2}w}{\partial y^2}\right)-Mw+f\left(x,y,t\right),$$

(61)

with the initial distribution and the boundary distributions:

$$w\left(x,y,0\right)=0,$$

(62)

$$w\left(\pm 1,y,t\right)=w\left(x,\pm 1,t\right)=0.$$

(63)

Define an exact solution for (61)–(63) as: $w\left(x,y,t\right)={\left(x-1\right)}^2{\left(x+1\right)}^2{\left(y-1\right)}^2{\left(y+1\right)}^2{t}^2$, the expression of the source term can be deduced:

$$\begin{array}{l}f\left(x,y,t\right)={\left(x-1\right)}^2{\left(x+1\right)}^2{\left(y-1\right)}^2{\left(y+1\right)}^{2}t\left(2+Mt\right)\\ -4\nu t^2\left(\frac{2\lambda }{\mathsf{\Gamma }\left(3-\alpha \right)}{t}^{-\alpha }+1\right)\left[\left(3x^2-1\right){\left(y-1\right)}^2{\left(y+1\right)}^2+\left(3y^2-1\right){\left(x-1\right)}^2{\left(x+1\right)}^2\right].\end{array}$$

(64)

Figure 2 presents the three-dimensional comparison behavior between the numerical and exact expressions. Obviously, the distribution of the numerical solution is basically the same as that of the exact solution, showing a bell-shaped curve that is high in the middle and low at both ends.

Table 1. The error and convergence order for space and the comparison of computational time between the finite difference scheme and the fast scheme when $\alpha =0.5$, $M=1$, $\nu =1$ and $\lambda =0.1$.

	Normal L1 Method			Fast Algorithm
$\mathit{\tau }=1/20000$	Error	Order	Time (s)	Error	Order	TIME (s)
$h_x=h_y=1/2^2$	$1.4745\times {10}^{-1}$			$1.4745\times {10}^{-1}$		6.58
$h_x=h_y=1/2^3$	$3.6738\times {10}^{-2}$	2.00	317.32	$3.6738\times {10}^{-2}$	2.00	8.78
$h_x=h_y=1/2^4$	$9.1903\times {10}^{-3}$	2.00	2737.48	$9.1903\times {10}^{-3}$	2.00	14.61
$h_x=h_y=1/2^5$	$2.3032\times {10}^{-3}$	2.00	16,424.92	$2.3032\times {10}^{-3}$	2.00	43.06
$h_x=h_y=1/2^6$	$5.8123\times {10}^{-4}$	1.99	59,840.35	$5.8123\times {10}^{-4}$	1.99	185.47

and

Table 2. The error and convergence order for time and the comparison of computational time between the finite difference scheme and the fast scheme when $\alpha =0.5$, $M=1$, $\nu =1$ and $\lambda =0.1$.

	Normal Scheme			Fast Scheme
${\mathit{h}}_{\mathit{x}}={\mathit{h}}_{\mathit{y}}=1/640$	Error	Order	Time (s)	Error	Order	Time (s)
$\tau =1/2^2$	$3.9307\times {10}^{-2}$		2749.74	$3.9307\times {10}^{-2}$		6.83
$\tau =1/2^3$	$1.9424\times {10}^{-2}$	1.02	2851.14	$1.9424\times {10}^{-2}$	1.02	14.90
$\tau =1/2^4$	$9.5638\times {10}^{-3}$	1.02	3057.22	$9.5638\times {10}^{-3}$	1.02	33.67
$\tau =1/2^5$	$4.7173\times {10}^{-3}$	1.02	3579.91	$4.7173\times {10}^{-3}$	1.02	70.72
$\tau =1/2^6$	$2.3347\times {10}^{-3}$	1.01	4622.72	$2.3347\times {10}^{-3}$	1.01	147.05

show the maximum error with the form $E\left(h_x,h_y,\tau \right)=\underset{0\le i\le M_x,0\le j\le M_y}{\max }\left|e_{i,j}^n\right|$, the convergence order for space with $r_s={\log }_2\frac{E\left(h_x,h_y,\tau \right)}{E\left(h_x/2,h_y/2,\tau \right)}$, for time with $r_t={\log }_2\frac{E\left(h_x,h_y,\tau \right)}{E\left(h_x,h_y,\tau /2\right)}$ and the computational time between the classical difference scheme and the fast scheme. The two tables show that the error is very small when verifying the accuracy of the numerical scheme and the accuracy is $\mathrm{O}\left(h_x^2+h_y^2+\tau \right)$, which is consistent with the analysis in the convergence in Theorem 3. Furthermore, the computational time indicates that the superiority of the fast scheme is that it can greatly reduce the calculation time without affecting the total accuracy.

Example 2.

The effects of the dynamic parameters on the distributions of velocity and shear force subject to various pressure with cosine forms.

Figure 3, Figure 4 and Figure 5 show the distribution of the velocity and shear force at $x=0$ (wall surface) with oscillating pressure gradient versus time with the form $-\frac{1}{\rho }\frac{\partial p}{\partial z}=\cos \left(t+1\right)$ when we choose $\nu =1$. The influences of the retardation time parameter on the velocity distributions and the distribution of shear force at the wall are shown in Figure 3. For $\lambda =0$, the influences of the retardation time disappear. With the appearance of the retardation time parameter, the big difference is that the overall distribution becomes lower with the physical, meaning that the retardation time parameter reflects a relaxation characteristic in slowing down the velocity propagation and decreasing the magnitude of the shear force at the wall. It can be concluded that a bigger the retardation time parameter corresponds to a larger the relaxation characteristic. The magnetic parameter has important impacts on the distributions of velocity and the shear force. The parameter $M=0$ indicates that the influence of the magnetic parameter is not considered. As shown in Figure 4, the consideration of the magnetic field makes the distribution at a fixed position smaller, and the value of the distribution becomes smaller when the magnetic parameter becomes bigger. The fractional parameter makes the velocity transport consider the memory characteristic. Figure 5 shows that the value of the distribution becomes smaller with an increase of fractional parameter.

The oscillatory frequency has important impacts on velocity distributions and the shear force distributions. Consider $-\frac{1}{\rho }\frac{\partial p}{\partial z}=\cos \left(wt+1\right)$, the three-dimensional velocity distributions and shear force distributions versus $y$ and $t$ with the effects of frequency are exhibited in Figure 6 and Figure 7, respectively. For $w=0$, the pressure is constant and the time parameter (for $t>0$) has no effects on the distributions. For $w\ne 0$, the distributions present as an oscillatory form and the bigger the frequency parameter is, the stronger the oscillatory character of the distributions will be. To discuss the effects of the various pressures with the space oscillatory flow, we consider $-\frac{1}{\rho }\frac{\partial p}{\partial z}=\cos \left(wz+1\right)$ with different $w$. The effects of frequency parameter on the velocity distributions and the shear force distributions versus $x$ and $z$ are respectively exhibited in Figure 8 and Figure 9. Similarly, the distribution curve shows that the distribution exhibits as a normal form for $w=0$. For $w\ne 0$, the distribution presents as an oscillatory form. Finally, the bigger the frequency parameter, the stronger is the oscillation of the distribution curve.

8. Conclusions

This paper considered the motion of fractional second-grade fluid in a straight rectangular duct. Both the analytical solution and the numerical solution were obtained. For faster computation, a fast scheme was proposed. Two examples were given. One illustrated the accuracy of the numerical solution and the advantage of the fast scheme. The other discussed the impacts of the involved parameters on the velocity distributions and the shear force at the wall surface. The results show that the retardation time parameter plays a role in a relaxation characteristic. The magnetic parameter and fractional parameter with the memory characteristic made the distribution of velocity and shear force become slower. The oscillation of the pressure versus space and time made the distribution present as an oscillatory form and for a larger frequency parameter, the oscillation of the distribution was stronger.